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alexgriva [62]
3 years ago
11

A 75-kilogram hockey player is skating across the

Physics
1 answer:
azamat3 years ago
6 0
I think it is (3) 690 N. You can use the theorem of Impulse and momentum where the impulse is equal to the change in momentum or:
F \Delta t=m v_{f}-m v_{i} (where f is for final and i for initial).
So you get:
F*0.65=0-(75*6)
F=692 N
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In the simulation, there are three balls on the floor. Drag each of them up off the floor, and then let go. See what happens to
Vlad1618 [11]

Answer:

I hope this helps and I'm not to late

A way the balls behave the same way is by bouncing about 1 time after throwing the balls up. A way the balls act differently is the blue ball is bouncier than all the balls, the red ball bounces about 2 times before stopping, and the green ball doesn’t really bounce except for one time.

Explanation:

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7 0
2 years ago
A student places his hand in front of a plane mirror as shown in the diagram. Which terms correctly describe the image that the
ANTONII [103]
Either A or D but maybe D
4 0
2 years ago
From the top of a tall building, a gun is fired. The bullet leaves the gun at a speed of 340 m/s, parallel to the ground. As the
Ivahew [28]

Answer:

The launching point is at a distance D = 962.2m and H = 39.2m

Explanation:

It would have been easier with the drawing. This problem is a projectile launching exercise, as they give us data after the window passes and the wall collides, let's calculate with this data the speeds at the point of contact with the window.

X axis

           x = Vox t

           t = x / vox

           t = 7.1 / 340

           t = 2.09 10-2 s

In this same time the height of the window fell

           Y = Voy t - ½ g t²

Let's calculate the initial vertical speed, this speed is in the window

           Voy = (Y + ½ g t²) / t

           Voy = [0.6 + ½ 9.8 (2.09 10⁻²)²] /2.09 10⁻² = 0.579 / 0.0209

            Voy = 27.7 m / s

We already have the speed at the point of contact with the window. Now let's calculate the distance (D) and height (H) to the launch point, for this we calculate the time it takes to get from the launch point to the window; at this point the vertical speed is Vy2 = 27.7 m / s

             Vy = Voy - gt₂

             Vy = 0 -g t₂

             t₂ = Vy / g

             t₂ = 27.7 / 9.8

             t₂ = 2.83 s

This is the time it also takes to travel the horizontal and vertical distance

            X = Vox t₂

            D = 340 2.83

            D = 962.2 m

           

            Y = Voy₂– ½ g t₂²

            Y = 0 - ½ g t2

            H = Y = - ½ 9.8 2.83 2

            H = 39.2 m

The launching point is at a distance D = 962.2m and H = 39.2m

6 0
3 years ago
define the term change and state one negative change you may encounter as a student or as an employee in the future​
Brums [2.3K]

Answer:

hey iam bored pls join

ID: 724 645 3790

Password: 1111

6 0
2 years ago
A 5kg ball is on top of the school building at a height of 40m above the ground.
mojhsa [17]

Answer:

A-Caclcuate the potential energy of the ball at that height

Explanation:

(a). Mass of the Body = 10 kg.

Height = 10 m.

Acceleration due to gravity = 9.8 m/s².

Using the Formula,Potential Energy = mgh

= 10 × 9.8 × 10 = 980 J.

(b). Now, By the law of the conservation of the Energy, Total amount of the energy of the system remains constant.

∴ Kinetic Energy before the body reaches the ground is equal to the Potential Energy at the height of 10 m.

∴ Kinetic Energy = 980 J.

(c). Kinetic Energy = 980 J.

Mass of the ball = 10 kg.

∵ K.E. = 1/2 × mv²

∴ 980 = 1/2 × 10 × v²

∴ v² = 980/5

⇒ v² = 196

∴ v = 14 m/s.

3 0
2 years ago
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